$A$ particle is executing linear $S.H.M.$ starting from the mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is (in $: 1$)

The mass of a particle is $1 \ kg$ and it is moving along the $x$-axis. The period of its oscillation is $\frac{\pi}{2} \ s$. Its potential energy at a displacement of $0.2 \ m$ is (in $J$)

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

The displacement-time graph of a particle executing $SHM$ is as shown in the figure. The corresponding graph between potential energy $(PE)$ and time is:

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

The equation of motion of a particle executing simple harmonic motion is given by $x=3 \sin \left(6 t+\frac{\pi}{6}\right)$,where $x$ is in metres and $t$ is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ is